The Henry's law constant fo the solubility of `N_(2)` gas in water at 298 K is `1.0xx10^(5)` atm. The mole fraction of `N_(2)` in air is 0.8. The number of moles of `N_(2)` from air dissolved in 10 moles of water at 298K and 5 atm pressue is

Options

A. `4.0xx10^(-4)`
B. `4.0xx10^(-5)`
C. `5.0xx10^(-4)`
D. `4.0xx10^(-6)`

Answer

A

Explanation

`P_(N_(2))=5xx0.8=4`
Applying Henry's law, we get
`P_(N_(2))=K_(H)xxx_(N_(2))`
`4=10^(5)x_(n_(2))`
`x_(n_(2))=4xx10^(-5)`
`x_(n_(2))=(n_(N_(2)))/(n_(N_(2))+h_(H_(2)O))`
`4xx10^(-5)=(n_(N_(2)))/(n_(N_(2))+10)`
`n_(H_(2))=4xx10^(-4)`